Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose you are given a nowhere-vanishing exact 2-form $B=dA$ on an open, connected domain $D\subset\mathbb{R}^3$. I'd like to think of $B$ as a magnetic field.

Consider the product $H(A)=A\wedge dA$. At least in the plasma physics literature, $H(A)$ is known as the magnetic helicity density.

How can one determine if there is a closed one-form $\mathbf{s}$ such that $H(A+\mathbf{s})$ is non-zero at all points in $D$?

The reason I am interested in this question is that if you can find such an $\mathbf{s}$, then $A+\mathbf{s}$ will define a contact structure on $D$ whose Reeb vector field gives the magnetic field lines. Thus, the question is closely related to the Hamiltonian structure of magnetic field line dynamics.

I'll elaborate on this last point a bit. If there is a vector potential $A$ such that $A\wedge dA$ is non-zero everywhere, then the distribution $\xi=\text{ker}(A)$ is nowhere integrable, meaning $\xi$ defines a contact structure on $D$ with a global contact 1-form $A$. The Reeb vector field of this contact structure relative to the contact form $A$ is the unique vector field $X$ that satisfies $A(X)=1$ and $\text{i}_XdA=0$. Using the standard volume form $\mu_o$, $dA$ can be expressed as $\text{i}_{\mathbf{B}}\mu_o$ for a unique divergence-free vector field $\mathbf{B}$. Thus, the second condition on the Reeb vector field can be expressed as $\mathbf{B}\times X=0$, which implies the integral curves of $X$ coincide with the magnetic field lines.

An example where $D=$3-ball and no $\mathbf{s}$ can exist:

Let $D$ consist of those points in $\mathbb{R}^3$ with $x^2+y^2 < a^2$ for a real number $a>1$. Note that all closed 1-forms are exact in this case. Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a smooth, non-decreasing function such that $f(r)=0$ for $r<1/10$ and $f(r)=1$ for $r\ge1/2$. Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be the polynomial $g(r)=1-3r+2r^2$. Define the 2-form $B$ using the divergence free vector field $\mathbf{B}(x,y,z)=f(\sqrt{x^2+y^2})e_\phi(x,y,z)+g(\sqrt{x^2+y^2})e_z$. Here $e_\phi$ is the azimuthal unit vector and $e_z$ is the $z$-directed unit vector. It is easy to verify that $B$, thus defined, is an exact 2-form that is nowhere vanishing.

Because $g(1)=0$ and $f(1)=1$, the circle, $C$, in the $z=0$-plane, $x^2+y^2=1$, is an integral curve for the vector field $\mathbf{B}$. I will use this fact to prove that the helicity density must have a zero for any choice of gauge. Let $A$ satisfy $dA=B$ and suppose $A\wedge B$ is non-zero at all points in $D$. Note that $A\wedge B=A(\mathbf{B})\mu_o$, meaning $h=A(\mathbf{B})$ is a nowhere vanishing function. Without loss of generality, I will assume $h>0$. Thus, the line integral $I=\oint_C h\frac{dl}{|\mathbf{B}|}$ satisfies $I>0$. But, by Stoke's theorem, $I=2\pi\int_0^1g(r)rdr=0$, as is readily verified by directly evaluating the integral. Thus, there can be no such $A$.

share|cite|improve this question
Can you explain this relation with the contact geometry and B-fields? – Chris Gerig Oct 10 '12 at 17:42
Just for the record, this was crossposted at MO as well:… – Neal Oct 14 '12 at 0:16

I believe that the answer, in generality, is that you cannot find $\mathbf s$. As an example, consider a field configuration that is translationally symmetric in, say, the $z$ direction (for example, if there is a smooth current density that directed parallel to the $z$ axis). Speaking in physics terms, if we assume that the helicity is non-zero in some gauge $\mathbf s=\boldsymbol\nabla \phi$, then without loss of generality, we can assume that $H > 0$ everywhere in the domain $D$. However, consider the line integral of $H$ over any closed field loop: $$0<\oint H\frac{\,dl}{|\mathbf B|}=\oint (\mathbf A + \boldsymbol\nabla\phi) \cdot\,d\mathbf l=\iint (\boldsymbol\nabla\times \mathbf A)\cdot\,d\boldsymbol\sigma= \iint \mathbf B\cdot \,d\boldsymbol\sigma=0$$ Here, $\,dl$ is a line measure, and $\,d\boldsymbol\sigma$ is the surface measure. The first integral is well defined since $\mathbf B$ is nowhere vanishing.

share|cite|improve this answer
You should be careful about defining $D$. In your example, $D$ cannot contain the the zeros of the field. If you suppose $D$ is $\{(x,y,z)|x^2+y^2>0\}$, then you will not be justified in changing your line integral into a surface integral using Stoke's theorem. In particular, suppose you start with a vector potential that is smoothly defined on all of $\mathbb{R}^3$ and then restrict it to this modified $D$. Now change gauge using $\lambda\nabla\phi$, where $\phi$ is the azimuthal angle and $\lambda$ is any constant. Then $\oint A+\lambda d\phi = 2\pi\lambda$. – Josh Burby Oct 13 '12 at 21:49
I managed to find a different example, similar to yours in spirit. See my edit. – Josh Burby Oct 14 '12 at 0:06
Unless there is a topological defect, the argument stands as it is. Just consider a current density $j(\rho)=\rho\exp(-1/|\rho-a|^2)$ in the domain $\rho>a$. Either way, I don't know of a general argument or condition pro or contrary to your question. – Ivan Oct 18 '12 at 18:51
The issue is $\mathbf{s}$ might not be the gradient of a scalar; it only need satisfy $\nabla\times\mathbf{s}=0$, which certainly has non-gradient solutions in your domain $\rho > a$. You won't come to the contradiction in your argument if you take into consideration these non-exact gauge changes. But thanks for your help. – Josh Burby Oct 19 '12 at 0:36
The example I gave allows extension to the entire space, which is unique if we fix the boundary conditions at infinity. Furthermore, under the same boundary conditions every $s$ such that $\nabla\times s=0$, has to be a gradient. Again, unless you have a topological defect (which can ruin your boundary conditions), you are pretty safe. – Ivan Oct 23 '12 at 20:14

Since $s$ is closed: \begin{eqnarray}H(A+s) &=& (A+s) \wedge d(A+s)\\ & =& (A+s) \wedge (dA + ds) = (A+s) \wedge dA \\ &=& (A \wedge dA) + (s \wedge dA)\end{eqnarray} If $dA$ is nowhere-vanishing on your domain, $|dA|$ should have a maximum value on the closure $\overline{D}$.

So you should be able to add a sufficiently large constant 1-form, so that $H$ is also nowhere vanishing.

share|cite|improve this answer
You're assuming $dA$ can be continuously extended to $\overline{D}$, which is not necessarily the case. Consider $A = \frac{1}{1-r} d\theta$ on the open unit disc (minus the origin). Then $dA = \frac{1}{(1-r)^2} dr\wedge d\theta$ which cannot be continuously extended to the boundary. (There may be physical considerations which imply every such physically possible $A$ can be extended continuously to the boundary - I have no idea). – Jason DeVito Oct 9 '12 at 17:26
(My example works in $\mathbb{R}^2$, not $\mathbb{R}^3$. To fix it, instead of $d\theta$, use $d\omega_{S^2}$, the area form on $S^2$.) – Jason DeVito Oct 9 '12 at 17:33
I don't think it's physical. To have magnetic field $H = dA$ be infinite, you must be very close to the magnetic pole. – cactus314 Oct 9 '12 at 18:12
There is a more serious issue than extending $dA$. Simply choosing $\mathbf{s}$ to be large and constant will not work in general. This is because $\mathbf{s}\wedge dA$ will typically have zeros for $\mathbf{s}$ constant. This is obvious if you identify $dA$ with a divergence-free vector field $\mathbf{B}$ using the standard volume form $\mu_o$ on $\mathbb{R}^3$. Then $\mathbf{s}\wedge dA=\mathbf{s}(\mathbf{B})\mu_o$. Near one of these zeros, the term $\mathbf{s}\wedge dA$ will not automatically dominate $A\wedge dA$, even if $|dA|$ is takes a maximum value on $D$. – Josh Burby Oct 13 '12 at 21:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.